IAO 2018, Теория, Група α — Задача 3. αβ-3. Залез на Слънцето в Коломбо

3.1.

We see in the picture that half of the solar disk disappeared over the horizon. This is exactly half of the disk with great accuracy, since we see its vertical edges strictly on the horizon. In zero approximation, this occurs at 18:00 local time. The accuracy of the fact, that the center of the Sun is on the horizon at the photo, can be estimated at $2'$ or ±8 seconds in units of time.

Let us consider what corrections to this time need to be taken into account and what is the accuracy of each the correction.

1. Longitudinal correction. The local time of Sri Lanka is UT+05:30, and it corresponds to the longitude $82^\circ30'$ East, the photographer was at the longitude $79^\circ51'$, i.e. $2^\circ39'$ to the west. This means that for him the sunset occurs $2^\circ39'\times4\ \mathrm{min}/{}^\circ=10\text{ min }36\text{ sec}$ later. The accuracy of this correction corresponds to $0.5'$ in longitude or 2 seconds in time.

2. Correction of the equation of time. According to the schedule, we find that on September 30 all events related to the Sun occur 10 minutes earlier. The accuracy of the correction is determined by the graph and is approximately half a minute.

3. The center of the Sun crosses the mathematical horizon strictly in the west and at 18:00 of true solar time only at the moments of equinoxes. The evening of September 30 in Colombo is 7 days and 11 hours later than the autumnal equinox. During this time, the Sun has moved along the ecliptic by about $7.4^\circ$. The declination of the Sun is approximately equal to $\delta=-23.44^\circ\times\sin7.4^\circ\approx-3.0^\circ$. Given the latitude of the place $6.9^\circ$, we obtain that the Sun is $3.0^\circ\times\sin6.9^\circ\approx0.36^\circ\approx22'$ lower than at the same time during the autumnal equinox. The rate of lowering the Sun is $15'/\mathrm{min}\times\cos\delta\times\cos\varphi$ ($\cos\delta$ arises due to the fact that the Sun moves not along a large circle, but along a small circle, and $\cos\varphi$ is due to the inclination of the trajectory). In our case these coefficients are very close to 1, so the Sun goes down $22'$ in 1 minute 28 seconds. That is, this correction is minus 1 minute 28 seconds. The correction accuracy is 4 seconds.

4. Refraction. The refraction near the horizon is $35'$. The Sun passes down this angle in 2 minutes and 20 seconds. This is a correction with a plus sign. Correction accuracy is 4 seconds.

5. Lowering of the horizon. It is said that the height of the camera above sea level was approximately $h=6.5\ \mathrm m$, with the radius of Earth $R\approx6400\ \mathrm{km}$, this gives lowering of the horizon for $(2h/R)^{1/2}\approx1.4\cdot10^{-3}\ \mathrm{rad}\approx5'$. This correction is plus more 20 seconds. The accuracy of the correction corresponds to 2 seconds in time.

In total, the sum all corrections is:

The largest error is given by the inaccuracy in determining the correction due to the equation of time on graph. That is, the event occurred at 18:01:48 with an accuracy of half a minute. Or at 18:02 if to write with a reasonable accuracy.

3.2.

Civil twilight ends when the center of the Sun appears $6^\circ$ below the mathematical horizon. As we calculated above, at the time of the shooting, the center of the Sun was at $35'+5'=40'$ and was moving downward with speed

The remaining $5^\circ20'$ before reaching $-6^\circ$ the Sun will pass in 21.5 minutes. That is, the answer to the second question of the problem is 21.5 minutes.

3.3.

The height of Adams Pk, as we see on the map, is equal to 2243 m. Lowering the horizon from this peak will be $(2h_A/R)^{1/2}\approx2.65\cdot10^{-2}\ \mathrm{rad}\approx1^\circ31'$. This is $1^\circ26'$ more than when photographing from the seafront. This effect raises the apparent position of the solar disk above the visible horizon (by $1^\circ26'$ compared to the camera on the seafront). On the other hand, measurements on the map (28 mm from the coast to the peak with a scale of 200 km to 40 mm) show that Adams Pk is 140 km East of Colombo, which corresponds to enlarging longitude by $360^\circ\times140\ \mathrm{km}/(40076\ \mathrm{km}\times\cos\varphi)=1^\circ15'$. The apparent position of the solar disk is lowered due to this effect. The difference of the effects is $11'$ or one third of the angular diameter of the Sun. Thus, from Adams Pk at this time the position of the Sun will be visible, at which 1/6 of its diameter is below the horizon, and 5/6 above the horizon. This picture should be drawn.

Basic criteria